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Chapter 6: Q. 07 (page 569)

A falling object accelerates downwards due to gravity at a rate of 9.8 meters per second squared. Suppose an object is dropped and falls to the ground from a height of 100 meters.

(a) Set up and solve a first-order initial-value problem whose solution is the velocity v(t) of the object at time t.

(b) Use your answer to part (a) to set up and solve a first-order initial-value problem whose solution is the position s(t) of the object at time t.

Short Answer

Expert verified

a) The differential equation:dvdt=-9.8, v(0)=0

The solution is:v(t)=-9.8t

b)The differential equation:dsdt=-9.8t,s(0)=100

The solution iss(t)=-4.9t2+100

Step by step solution

01

Step 1. Given information

The ball is dropped from height of 100 m.

The acceleration due to gravity is9.8m/s2

02

Part(a). Step 1.  Set up and solve a first-order initial-value problem whose solution is the velocity v(t) of the object at time t. 

The differential equation is:

dvdt=-9.8, v(0)=0

localid="1652102005142" dvdt=-9.8dv=-9.8dt鈭�dv=-鈭�9.8dtv=-9.8t+c

When v(0)=0

Then

localid="1652102017053" 0=-9.8(0)+cc=0

So the solution to the differential equation is:

v(t)=-9.8t

The acceleration has a negative sign since the object falls downward.

03

Part(b). Step 1. To set up and solve a first-order initial-value problem whose solution is the position s(t) of the object at time t by using the answer of part (a).

The differential equation of position:

dsdt=vdsdt=-9.8t

The initial height is 100 m.

So, s(0)=100

Now solve the differential equation:

dsdt=9.8tds=-9.8tdt鈭�ds=鈭�-9.8tdts=-9.8t22+cs=-4.9t2+c

Since, s(0)=100

Hence,

100=4.9(0)+cc=100

So the solution of the differential equation is:

s(t)=-4.9t2+100

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